Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
#define rep(i, a, n) for(int i = a; i <= n; i++)
const int maxN = 3010;
int n;
double p[maxN], dp[maxN][maxN];
int main()
{
	cin >> n;
	rep(i, 1, n) cin >> p[i];
	dp[0][0] = 1.0;
	rep(i, 1, n) rep(j, 0, i) dp[i][j] = dp[i - 1][j] * (1.0 - p[i]) + dp[i - 1][j - 1] * p[i];
	double ans = 0.0;
	rep(i, (n + 1) / 2, n) ans += dp[n][i];
	printf("%.12f\n", ans);
	return 0;
}